Question

A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sample of 46 bins, the sample mean amount was 49.41 pounds and the sample standard deviation was 3.99 pounds. Construct a 92.7% confidence interval for the expected amount of garbage per bin for all bins in the city. Answer to 3 decimals (a) What is the lower limit of the 92.7% interval

Answer 1

Answer:

The 92.7% confidence interval for the expected amount of garbage per bin for all bins in the city is between 48.135 pounds and 50.685 pounds. The lower limit is 48.135 pounds.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 46 - 1 = 45

92.7% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 45 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.927}{2} = 0.9635$. So we have T = 2.1437

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.1437\frac{3.99}{\sqrt{45}} = 1.275$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 49.41 - 1.275 = 48.135 pounds

The upper end of the interval is the sample mean added to M. So it is 49.41 + 1.275 = 50.685 pounds.

The 92.7% confidence interval for the expected amount of garbage per bin for all bins in the city is between 48.135 pounds and 50.685 pounds. The lower limit is 48.135 pounds.